2014 AMC 10A Problems/Problem 8: Difference between revisions
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In order for <math>\frac{n+1}{2}</math> to be a perfect square, <math>n+1</math> must be twice a perfect square. From the answer choices, <math>n+1=18</math> works, thus, <math>n=17</math> and our desired answer is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math> | In order for <math>\frac{n+1}{2}</math> to be a perfect square, <math>n+1</math> must be twice a perfect square. From the answer choices, <math>n+1=18</math> works, thus, <math>n=17</math> and our desired answer is <math>\boxed{\textbf{(D)}\ \frac{17!18!}{2}}</math> | ||
==See Also== | ==See Also== | ||
Revision as of 13:27, 26 January 2015
Problem
Which of the following number is a perfect square?
Solution
Note that for all positive 
, we have
![\[\dfrac{n!(n+1)!}{2}\]](http://latex.artofproblemsolving.com/c/6/c/c6cadcea118b402a6a8d08b9ed33371afc4b43a0.png)
![\[\implies\dfrac{(n!)^2\cdot(n+1)}{2}\]](http://latex.artofproblemsolving.com/0/f/4/0f4a73e6e6c4cc24be518527f91d6a867e8f5b24.png)
![\[\implies (n!)^2\cdot\dfrac{n+1}{2}\]](http://latex.artofproblemsolving.com/2/1/3/2130d4422339b6c4b17ef79fc45d476fffa67ed2.png)
We must find a value of such that 
is a perfect square. Since 
is a perfect square, we must also have 
be a perfect square.
In order for to be a perfect square, 
must be twice a perfect square. From the answer choices, 
works, thus, 
and our desired answer is 
See Also
| 2014 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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